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12:00 AM
You wrote partial derivatives (and do NOT ever write something like $\partial f/\partial g$ or $df/dg$ — it is meaningless).
 
i like to think of tangent planes to the graph
not a particularly unusual perspective, of course
 
How does that elucidate the composition of linearizations?
 
it does not.
 
Gosh ... This might be the second time today I am being analytic and ungeometric. I really will lose my geometer's license.
 
i am arriving late as usual
 
12:03 AM
composition of linear functions multiplies slopes
YES! thank you!
 
In one dimension, yes.
 
right right
 
Glad that helped your intuition :)
 
 
2 hours later…
2:08 AM
asked the daughter what her favorite activity of the day was. expected to hear about the pony again, instead got "i liked to watch the birds hopping around." on our porch, which she sees every day. not the trip to ride a pony, or the trip to the park to swing on swings. such a toddler.
 
Good Morning!
 
If there are people seeing errors in my solutioncan they please comment (at least)?
@leslietownes Hello!
Thank you.
 
2:38 AM
never mind
 
2:55 AM
yo
 
I remember a while ago seeing a picture of a pattern on the hyperbolic plane that looked like it took up 1/2 of it, then when you shifted your view it looked like 1/3 of it instead. I think it was in a question here but I'm not 100% sure. Does anyone know what/where this is?
 
@TedShifrin I think I recall why I kept avoiding Taylor. I can see the utility, but they are super tedious to write out (and with all the sign flipping in cos / sine, a real headache for my dyslexia)
but the expansion of e^x is nice
 
3:19 AM
so, is there any rigorous definition of quotients of differentials?
 
several of the answers to mathoverflow.net/questions/73492/… are thoughtful and somewhat relevant
 
so that they act independently, and aren't treated as a symbol for a limit
thanks, will look into that!
 
i like steven landsburg's answer, which characterizes treating dy/dx as a fraction as a 'gateway drug' to more harmful practices. we may not always agree politically but we agree on that.
 
I guess that if you make sure you can always subsitute $dx$ or any differential by $x - x_0$ and add in an error function that goes to zero, you can safely treat it as a real ratio
 
4:27 AM
may the math gods bless Ted Shifrin and the math 3500 lectures
 
4:46 AM
yo
mathics is also nice in plotting
 
5:06 AM
scilab is also nice and it has many features
idk why i am telling this here lol
 
 
2 hours later…
7:20 AM
anyone knows how the author goes from this, to:
$f(x_0 + h) - f(x_0) = f'(x_0)h + \epsilon h$
where $h = x - x_0$
I tried limit algebra, and I died
 
shintuku, that follows from definition of limit.
 
I thought about that, but how are they getting an equality from a limit?
so, of course, the initial equation can be written $\lim \limits_{h \to 0} \text{blabla} = 0$
then, by limit algebra, you get something like $\lim \limits_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = \lim \limits_{h \to 0} f'(x_0)h$
but where the hell is the $\epsilon h$ coming from
is this a canadian limit of some sort
in any case, I've been at this for two hours, I beg for help
 
Think about this: if $\lim f(x)=2$ then $f$ can be defined as $f(x)=2+g(x)$ where $\lim g(x)=0$
 
wait, this follows from the epsilon-delta definition?
 
So your $\epsilon$ is “something” that tends to $0$ as $h\to 0$.
@shintuku you can prove that using $\epsilon$ delta also
 
7:32 AM
$f(x) = 2 + g(x)$ would be some function that gets away from 2. Right. and as $g(x)$ as a function of $x$ approaches whatever value it is that makes $f(x) = 2$, then $g(x) = 0$
 
$g(x)=f(x)-2$ so what happens to $\lim g(x)$?
 
my bad, are we assuming $x$ goes to 0?
nevermind, that's irrelevant
 
In my example, it doesn’t matter shintuku. Do you see why?
If this is clear, use similar arguments in your question about derivative.
 
ah, I got it
niiiiice
thank you!
if $\lim f(x) = 2$ holds, then for $f(x) = 2 + g(x)$, whatever $g(x)$ is, it will be $0$ when $x$ goes to whatever it goes to with $\lim f(x)$
 
Great!! 😊
 
7:47 AM
wow, finally
I've been at this for too long
thanks a lot
 
Morning exercise: Try to prove limit rules using this (by introducing g(x) ) See how simple the proofs become
welcome :)
 
8:36 AM
@Koro quick follow up, if you're up for it (already greatly appreciate the last answer). I've proven the property you mentioned: if $\lim \limits_{x \to x_0} f(x) = L$ exists, assume $| x - x_0| < \delta$, then $-\epsilon < f(x) - L < \epsilon$, which means $f(x) + g(x) - L = 0$ exists, so $f(x) = L - g(x)$ if $|x - x_0| < \delta$. but, would you happen to know why we'd use $\epsilon h$ as a $g(x)$?
looks to me like we can use literally any $g(x)$, as long as $\lim \limits_{x \to x_0} g(x) = 0$
 
8:57 AM
Well, in this case please note that we are taking $\epsilon $ as $g(h)$ and not $\epsilon h$.
Also note that $\epsilon h$ in this case can be replaced by $o(h)$.
 
9:20 AM
I'll look into that, thanks!
 
10:04 AM
You know how we read C(m,n) as "m choose n"? Is there a similar way to read P(m,n)?
 
I need guidance with this inequality. consider $|a|·|b|\lt\varepsilon$ where $\varepsilon$ is arbitarily small. If $0\lt|b|\lt10$ then will this $|a|·10\lt\varepsilon$ inequality hold true ?
 
 
3 hours later…
1:24 PM
@sayanGhosh Thanks! I had shared the questions here before: chat.stackexchange.com/transcript/message/57924025#57924025
I had missed your messages earlier, sorry about that.
 
Let $M$ is a finitely generated $A$-module where $A$ is a local ring with maximal ideal $m$. Then $M/mM$ is a $A/m$-vector space so there is a basis $x_1+mM,...,x_n+mM$ of $M/mM$. Now let $L$ be a free $A$-module generated by $x_1,...,x_n$. Define $\phi:L\to M/mM$ by $x_i\mapsto x_i+mM$. Then I want to show the kernel $\{\sum_{i=1}^na_ix_i| \sum_{i=1}^na_ix_i \in mM\}=mL$. $\supset$ is clear but how can I prove the reverse inclusion?
 
1:56 PM
cellular automation
I got problem with it
anyway I am too tired to discuss about it
 
2:16 PM
@PeterJohn the induced map $L/\mathfrak{m}L\rightarrow M/\mathfrak{m}M$ is an iso, so the kernel can't be bigger
 
2:41 PM
@Thorgott Oh, thank you
 
2:53 PM
@BalarkaSen Yeah. Got it. Btw, if you wish, can I talk to you through email or via discord bcoz I'm also interested in Differential Geometry and in Algebraic Topology (Homology, Cohomology) as yours afaik.
 
3:19 PM
Hi, how would you show the countable sub-additivity of the Hausdorff measure? The book I'm reading said it's a particularly tough task
 
3:33 PM
@sayanGhosh Yeah you can send me a message on discord
 
4:06 PM
How come if $\det(ABC)=\det(A)\cdot \det(B)\cdot \det(C)$ I can't commute the $\det$'s in the RHS around and obtain arbitrary permutations of $(ABC)$ have the same determinant, unless it's true that a product of matrices has the same determinant regardless of order?
Perhaps the $\det$ function doesn't commute with itself?
 
It's true @Charlie
 
you can
 
det(AB) = det(BA) = det(A)det(B)
 
oh
oh yeah of course I can see how that generalises, ty
 
Can someone help me here?
H^s(F) is the s-dimensional Hausdorff measure
but I don't understand their explanation
 
4:32 PM
@BalarkaSen Thank you! Please share the details of your account.
 
which part is the issue
 
4:48 PM
@Thorgott So I see how the result follows if the Hausdorff measure is zero for all s > n
I just need help with proving that the Hausdorff measure of bounded sets is zero
After that I can use countable sub-additivity of the Hausdorff measure to conclude the same for all sets in R^n
"By considering covering of \delta mesh cubes, ....., H^s(F) = 0"
this part is the issue
 
H^s of an n-dimensional cube is 0 if s>n
might be annoying, but this can be computed from the definition
 
the calculation seems a little crazy
oh but yes that will do the trick
@Thorgott ok so i want delta-covers for the cube --- will it be okay to only consider smaller cubes covering this one?
 
you just need to partition a cube into smaller and smaller cubes
 
5:06 PM
i think our definitions of H^s(F) may be different
 
tell me yours
 
H^s(F) is lim_{\delta\to 0} H^s_\delta(F)
|U_i| = diameter of U_i
\delta-cover means that every set has a diameter \le \delta
*every set in the cover
 
that's my definition too
partitioning a cube into small cubes gives you upper bounds for $H_{L/n}^s$ that go to $0$ as $n\rightarrow\infty$ (here, $L$ is the side length of the cube)
 
then why do you think that only cube covers work?
@Thorgott what's L?
and is n the same n as in R^n
 
no, ns just a variable
 
5:11 PM
refers to number of smaller cubes?
 
no
 
then? :(
oh okay
so i've to find smaller cubes of diameter atmost L/n
 
just partition a cube into smaller cubes and see what you get
the L/n I just chose to get actual partitions, but this isn't a significant point
 
@Thorgott mathb.in/55241
This is what I get
doesn't really go to zero as n goes to infty unless I've missed something
 
you forgot taking to the $s$-th power in the RHS
 
5:23 PM
yup i've edited that
 
now what happens as $m\rightarrow\infty$
 
the RHS goes to zero!!!!
and delta goes to zero
wow we did it
thanks!
 
For a result that people often call elegant, Conway's actual proof on the look and say sequence is surprisingly computationally intensive link.springer.com/content/pdf/10.1007/978-1-4612-4808-8_53.pdf
I mean knowing that the end result is an algebraic number of degree 71 might give you a clue beforehand, but it was still worse than expected
 
np, glad you got it
 
@TedShifrin Ooooooh. I read that book. And I really liked it.
 
5:33 PM
derivative, what is the statement proved in that paper? it's paywalled
 
@sayanGhosh Send me an email on my university email and I'll share it there
You know my ISI email right?
 
oh, i found a non-paywalled version
 
call the lengths of the terms in the look and say sequence $L_i$. Then $L_{n+1}/L_n\to\alpha$ where $\alpha$ is..
oh
weird, I'm on my home computer, idk why it isn't paywalled for me, but maybe it's because I accessed my university's VPN on it recently
 
Here's a dumb question. I have an ODE $u''(z) + p(z) u'(z) + q(z) u(z) = 0$ where $p, q$ are meromorphic functions on $\Bbb C$. Somehow, $z = 0$ being a "regular singularity" is the only interesting possibility, in which case $p$ has a pole of order $1$, and $q$ has a pole of order $2$ at $z = 0$. Why?
 
my guess is that it seems elegant because it cleanly cabins the computational aspect away from the concepts, which are simpler and very well tuned to the problem.
it's maybe no coincidence that i found a copy of the paper on one of doron zeilberger's course website. a lot of his work is like that too even if it is frightfully computational.
 
5:38 PM
There is a classical theorem of Fuchs which describes all the solutions when you have such an ODE with a regular singularity.
I guess if you write $v(z) = u'(z)$ then you have the system $(u', v') = A(u, v)$ where $A(z) = (0, 1|-q(z), -p(z))$
So now we can run some monodromy business. I forget how this works
Locally this will always have solutions away from $z = 0$, the problem is to patch, which requires passing to some cover
 
@BalarkaSen Yes. Sure.
Done.
 
5:56 PM
Sent @sayanGhosh
 
P.s: I know that since your very childhood you are keen to Mathematics and you are a wonderful Mathematician as I believe. So, I'm very much eager to talk to you 🙂
 
I'm always around and happy to talk about math. But we have met before, haven't we?
 
Surely not. But I heard about you and your passion for Mathematics few years back from your relative, I think.
 
Oh, I might have confused you with someone else, then. Where do you study?
I wouldn't put my interest in mathematics in lofty terms. It's a good way to pass time, just like playing a good video game
Most of my relatives neither understand video games nor academia so they have a broken perspective of how things are
 
About the opposite inequality, I'm not sure how the author gets to it
The fact that the s-dimensional Hausdorff measure of the union is 0, only tells us that s > \dim_H \bigcup_{i=1}^\infty F_i
and we have assumed s > \dim_H F_i for every i
but this doesn't allow us to compare \dim_H F_i and \dim_H \bigcup_i F_i right?
 
6:08 PM
The last sentence tells you the Hausdorff dimension of $\bigcup_{i = 1}^\infty F_i$ is less than any number bigger than supremum of the Hausdorff dimensions of each of $F_i$
 
dave renfro's comment in math.stackexchange.com/questions/3827302/… might help
or balarka's comment right here :)
 
So $\dim_H(\bigcup_{i = 1}^\infty F_i) \leq \sup_i \dim_H(F_i)$, no?
Thanks for verifying, leslie
 
it's using the characterization of sup as smallest upper bound
 
@BalarkaSen No wait, how?
 
Fix any $s > \sup_i \dim_H(F_i)$. Then $s > \dim_H(F_i)$ for all $i$.
So $\dim_H(\bigcup_i F_i) < s$ by what you wrote.
 
6:12 PM
@BalarkaSen Do you want to fix s = \sup_i ?
 
Not really.
 
@BalarkaSen Agreed but how does this help us compare \dim_H(\bigcup_i F_i) < s and \sup_i \dim_H F_i? both are less than s, that's all I know
 
Hi Everyone :) What's the common interpretation of X^+ and X^- where X is a series?
 
@epsilon-emperor By what I said, you have $\dim_H(\cup_i F_i) < s$ for all $s > \sup_i \dim_H(F_i)$. If $a < s$ for all $s > b$, then $a \leq b$.
 
@BalarkaSen It might be. Great
 
6:15 PM
wklm: if a_n is the sequence whose partial sums make up the series, a_n^{+} might be the sequence taking the value a_n, if a_n >= 0, and 0 otherwise. similarly a_n^{+} might be the sequence taking the value 0 if a_n >= 0 and a_n (or perhaps even -a_n) otherwise
 
@sayanGhosh Ah, this makes sense now. Thank you!
 
so you end up having a_n = a_n^{+} plus or minus a_n^{-} and have notation for just collecting positive and negative terms
 
How would you prove that statement about a, b and s? Sorry it's probably very elementary
 
sometimes see this in the proof of riemann's theorem on rearrangements of partial sums of conditionally convergent series
 
It's intuitive but I'm using it for the first time somehow
 
6:17 PM
@epsilon-emperor Hint: Take inspiration from your username.
OK, better hint. Prove by contradiction.
:)
 
@BalarkaSen Done! With inspiration from username XD
 
thanks a lot @leslietownes
 
Suppose a > b. Then there exists \epsilon > 0 such that a > a- \epsilon > b. Our hypothesis gives a < a - \epsilon, contradiction! :)
Thank you!
 
@RyanUnger Do you agree that pseudoholomorphic curves should be thought of as quasi-minimal surfaces, where quasi = in the Gromov sense?
 
@Thorgott hmm we didn't use that in Balarka's proof
 
6:30 PM
@BalarkaSen OK, I think the point is if $\mathbf{u}(z) = (u(z), v(z))$, then $$\mathbf{u}'(z) = \sum_{i \geq 1} \frac{A_i(z) u_i(z)}{z - p_i}$$ where $A_i(z)$ are holomorphic and $p_1, \cdots, p_n$ are the regular singularities.
In this sense, hypergeometric equations are kind of universal because they solve the Riemann-Hilbert problem for all possible representations $F_2 \to \text{GL}_2(\Bbb C)$. OK.
 
its implicit
 
@Thorgott Okay cool I'll think about it. Thanks!
 
7:02 PM
Howdy @Thor and a @Balarka
Hi epsilon and negative @Leslie.
 
hey Ted
 
good afternoon
 
happy drizzly afternoon, @Leslie.
 
it's only cloudy here. although we could use some drizzle it would cut short the planned afternoon trip to the duck pond. so i hope it stays cloudy.
 
It drizzled/rained on me for my 2 1/2 hours going to the Farmers Market and other shopping.
 
7:09 PM
great timing!
 
Yup. But you know that ducks love rain :)
 
7:47 PM
lovely ride through tilden to alvarado and back via kensington. super windy at the crest.
back to defcon 5
 
the PSQs are safe again
 
ROFL
 
8:05 PM
i did sneak a look at a few :-)
 
8:17 PM
This is far from a PSQ, but you both might be interested. I've never seen anything like this.
 
yeah, wow. i've seen some goofy expressions for the taylor remainder but nothing like that. very weird to see p > 0 in something like this when 1 is a more common dividing point for introducing p.
 
You can either round numbers, but you can also just discard decimals
What's the exact term for that latter thing again?
not concatenate...
i just can't remember the exact word!!
truncate!
thx guys you are so useful and great
 
you are very welcome.
if it's true presumably it pops out of applying the mean value theorem for integrals to some exact expression for the remainder involving one or more integrals. i'm trying to imagine a purpose of this. i was thinking an embedding of a function space defined by norms involving differentiation into something simpler, or a renorming of a function space with something simpler.
it drives me crazy when people don't say where the thing came from. i never thought i'd say this, but would it kill him to post a cell phone photo of where he saw it in print?
 
8:34 PM
morning
speaking of Taylor, I think I am 90% of the way to prooving Eulers formula
I realised that half of the i's in e^ix become -1's and the other half become -i's
 
around 2001 we had a mystery pdf of something we wanted to cite it if its proof resembled ours (we only had a few pages with a statement of the result and the first two lines of a proof). the guy who found it couldn't remember where he'd gotten it and google didn't index pdfs at the time. we were not able to figure out where it came from.
about 15 years later i googled it and it was from a partially completed chapter of something one of his coauthors had drafted up but never completed. the proof was not complete.
@andrew exactly.
 
now I'm just scratching my head at what Euler was doing to have noticed this formula in the first place :P
 
Hello everyone, I was wondering whether anyone knows how to test if an exponential distribution is a suitable model for the quakes data in R.
 
like a goodness of fit?
the folks over at stats.stackexchange.com might know
 
@AndrewMicallef Have you seen this answer?
 
8:40 PM
is it about euler (i still havent got my proof finished, just thought that was a profitable observation so if it is eular I will decline to click that link)
 
However, plugging $ix$ into the series for $e^x$ and comparing to the series for $\sin(x)$ and $\cos(x)$ is pretty simple.
 
at least in that book, he derives it from the series, as you did.
 
@AndrewMicallef okay, then don't click that link. However, it is a completely different approach than what you are doing.
 
it's written in an old enough style that it may not be much help :)
"After logarithms and exponential quantities have been considered, circular arcs and the sines and cosines of these must be considered; not only because they constitute another kind of transcending quantity, but also because of the logarithms and exponentials of these that arise when they are involved with imaginary quantities,
which will become clearer below." you said it, euler.
 
@robjohn no I hadn't seen that and it is totally different to the approach I was taking. Ted sent me off on Taylor series....
 
8:44 PM
which is not a bad way if you can use Taylor Series
 
also side bar before I run off, any chance chatjax will work on firefox for android in the near future?
@leslietownes yeah I find modern maths hard enough to read
 
@AndrewMicallef It doesn't? I thought it did. If not, have you tried the suggestion for Chrome on Android?
 
I'll give chrome a look...let you know later today when I have my break
 
I thought there were people here who were using ChatJax running Firefox on Android.
 
yeah on firefox when I do the bookmark trick it just loads infinitly then craps out (not the best description, but i'm pulling my memory for the symptoms)
 
8:47 PM
You are running Firefox on an Android phone?
 
yeah
maybe now (5 mins before I have to run) wasn't the best time to bring it up :p
 
okay. See if the suggestion for Chrome on Android helps with Firefox
 
will do
 
euler is easier to read than a lot of math of similar vintage but it is not as easy to read as at least some modern books.
that should be the test of a real analysis book. is it easier to follow than euler.
 
Can anyone help me with a financial mathematics question regarding Wiener ProcesseS?
 
9:00 PM
@TedShifrin very odd Taylor expression. I think I would like the OP to provide more background before I would spend time on that one.
 
9:26 PM
copper, in comments he says he was 'checking some documents with taylor series' and found that formula. what more context do you want
this was after, not before, he was asked for a source
i'm going to turn this into a best selling novel where the source of the problem is tracked down after deciphering a series of clues in a number of international locations. i'm thinking of calling it The Taylor Code
coming soon to airport bookshops near you
 
To all my haters in this room:
I hope you lose your watch and are late for an important appointment.
 
oooof 🔥
 
9:44 PM
what's a watch? that quip must predate the widespread use of mobile phones.
like "here's a [dime, quarter], call someone who cares."
here's my samsung galaxy s5, call someone who cares
 
10:06 PM
Hi everybody
Hi Leslie, Ted, Robjohn
Hi copperhat
I have a question on continuity
related to "shadow points"
It is to be proven that for all $\displaystyle x\in \ ( a,b) ,$ we must have $\displaystyle f( x) \leq f( b)$
My question is: Is continuity really required for proving $f(x)\leq f(b)$?
I think that intermediate value property possession by $f$ is enough.
Am I missing something? Thanks.
 
@leslie @copper Here!! Whew! I got it.
 
wow. tiny tex hiccup there, something isn't typesetting.
 
@Koro: You might want to tell people (other than me, since I know the problem well) what the actual question is. I.e., what is a shadow point and what is the hypothesis?
@leslie to Koro or to me?
 
to you, after "and now it all falls in place."
 
Oh, weird. It was OK and then it isn't. Let me check.
 
10:15 PM
i read somewhere that the preview engine is based on a slightly different flavor of mathjax than the display version once a question is submitted. sometimes the two do not coincide. i ran into this a lot a long time ago. maybe that is still the case.
 
@Ted: I saw your comment there and I also tried to solve the Taylor series problem. I got $(x-a)^{p+1}$ and $p$ also but didn't get $(x-a)^p$ 😁
@Ted: I have incorporated the question in Latex version in the link I have shared above.
 
OK, @leslie. Thanks.
It's fixed. I had so many braces I messed up a few.
 
typesetting is fine now.
Any hint/suggestion for my question please?
 
@Koro: Don't be impatient.
Oh, you have an older version of Spivak's book. The problem was revised because I pointed out to him that one of my wise students had a far easier proof.
 
Ahh, I see.
 
10:23 PM
Sometimes it isn't useful to state an arcane hypothesis if in reality it gets you nothing extra. I mean, the examples of functions that are discontinuous and have the IVP are pretty limited. However, here we want to show that $f(a)=f(b)$. Without continuity, how are you going to get $f(a)\le f(b)$?
 
are they pretty limited?
I'd think discontinuity is the more generic phenomenon, even amongst Darboux functions
 
I'm afraid till part (a), there is no mention of showing $f(a)=f(b)$
Although that comes in later part of the same question.
 
more generic can still be arcane. :)
in fact, quasi-all functions are arcane. the proof uses the baire category theorem
 
@Koro It's not always optimal when writing a problem to give something too general for part a, and then make it more specific for part b. The way the problem got rewritten, it says "Suppose $f(a)>f(b)$. Show that the point where $f$ takes on its maximum on [a,b]$ must be $a$." Without continuity we wouldn't even know there's a max.
@Leslie: There was still a TeX error in the proof (a missing fraction), but I have fixed it. I think it's done.
 
Ok. I'll think more about the problem.
 
10:28 PM
Once I remembered Lagrange's trick of varying the point $a$, it wasn't too hard to see how this should go. I wonder why this generalization of the MVT is useful.
@Koro I no longer have the older editions with that statement of the problem. My recollection is that one of my students, having done the exercise, said something like: "Why do all this sup/inf stuff? Just show directly that ..."
 
Okay. Noted.
 
So I no longer remember the original statement that you're attached to.
 
10:44 PM
@TedShifrin Very nice!
the pointless harassment continues: "Does this answer abide by the EoQS, which you've previously been linked to?. No, it doesn't"
 
Why are they so hot on your tail, @copper?
 
Danilo's explanation of the application of that result is not illuminating.
i want a photo of the secret document
 
it is just one person who seems to track down my 'transgressions'
not quite sure i get the point of tracking down such stuff on questions that have already been closed.
and a 'punitive' downvote. like i give a flying whatever
 
Have you gone into the mod room to engage in a quiet conversation about this, @copper?
 
i added a comment to cured
 
10:54 PM
@Leslie: I just added my two cents' worth about arcane for more arcane.
 
there's a post on meta which indicates that starting with one post as a jumping off point to browsing through someone's history and tracking down other transgressions is discouraged. at least one mod feels very strongly about this.
 
@leslietownes I would not wish my biggest enemy to lose his smartphone. That's too cruel. How is he supposed to check his facebook feed while on the bus otherwise?
 
you are a greater humanitarian than i am.
 
our dependence on phones is disturbing
 
@leslietownes Such kind words... You deserve to keep your watch
 
11:05 PM
@Leslie Have we ducked?
 
pretty neat answer @TedShifrin
 
Thanks, copper. Not a PSQ.
 
:-). i was going to say that technically it is a psq :-)
 
Then you should have answered it!
 
beyond my abilities.
i need some sort of context usually to even start...
 
11:11 PM
I stole from Lagrange.
I just want one application.
 
i love when the old tricks still work. i just wish there had been a reason for it.
 
Good artists copy; great artists steal
 
Spivak has precisely one application of the Cauchy remainder formula, of which this is a generalization. He said he knew of no other.
Maybe Terry Tao knows of one.
 
I did not realise that Erdos was a tutor of Tao
 
Nor I.
 
11:17 PM
that's what kept me from the fields medal, i think. no erdos.
 
Rather orthogonal to your own interests.
 
i would've had different interests! i'd be doing graph theory. and amphetamines.
 
Number theory.
 
and now i'm too old to qualify. that's OK, it's time to make room for the next generation. i don't need a spotlight.
 
You burned them all out.
 

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